0
$\begingroup$

Given sequences $\{a_n\}$ , $\{b_n\}$ of positive real numbers and $\textrm{ limsup } \frac{a_n}{b_n} < \infty$. Prove there is a constant M such that $a_n \leq Mb_n$

Defn: Let $(a_{n})_{n=1}^{\infty}$ be a bounded sequence. Define the sequence $c_n = \sup \{a_{k}: k \geq n\}$ for $n \geq 1$. If the sequence $c_n$ converges, then the value it converges to is the limit superior of $(a_n)$.

Attempt:

So given that the lim sup is finite, it means there exists a value M such that for all $n > k$, $$\frac{a_n}{b_n} < M \\ \Rightarrow \ a_n \leq Mb_n$$

Comment: Surely there is more to it than this and I have missed something.....

$\endgroup$

marked as duplicate by Jeff, YuiTo Cheng, воитель, Thomas Shelby, Leucippus Jun 20 at 5:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ You haven't been explicit about where the $M$ comes from. Is it something like $\limsup + \epsilon$? And that only works for $n>k$. Your original problem was stated for all $n\in\Bbb N$. $\endgroup$ – Ted Shifrin Jun 19 at 23:43
  • $\begingroup$ Still amazed that these legends of Mathematics contribute to my simple questions...........The question didn't explicitly state anything about the $M$. All I've done up to this point is prove the usual properties of $\limsup$ and $\liminf$ that most analysis textbooks have you start out with. $\endgroup$ – dc3rd Jun 19 at 23:51
  • 1
    $\begingroup$ Well, then write it out carefully please. I don't see (without a little work) where your sentence comes from, and I added my additional complaint/suggestion. $\endgroup$ – Ted Shifrin Jun 19 at 23:52
  • $\begingroup$ What is the definition of $\limsup$ that you are working with? $\endgroup$ – José Carlos Santos Jun 19 at 23:54
  • $\begingroup$ What specifically did you mean to write out carefully? My assumption on the $limsup being finite? @JoséCarlosSantos just edited the question. $\endgroup$ – dc3rd Jun 19 at 23:58
2
$\begingroup$

It's not quite as simple as you make it in your proof because the sequence may approach the $\lim \sup$ from above. The following proof works, though:

Let $M_1= \lim \sup \frac{a_n}{b_n}$. Then $\exists N~ n \gt N \Rightarrow \frac{a_n}{b_n} \lt M_1+1.$ Let $M_2 = \max \{ \frac{a_n}{b_n} \vert ~n \leq N \}.$ Let $M = \max \{M_1+1, M_2+1 \}$. Then $\forall n \in \Bbb N~\frac{a_n}{b_n} \lt M \text{ so } a_n \lt Mb_n.$

$\endgroup$
  • $\begingroup$ In your comment you mentioned that the sequence $b_n$ is monotonically decreasing. So you defining $M_2$ as you did accounts for the possibility of there being a larger value than the $\limsup$ from earlier in the sequence. $\endgroup$ – dc3rd Jun 20 at 0:47
  • $\begingroup$ I meant $b_n$ in the definition you've given of $\lim \sup$. It's a little confusing because you used the same letter as in your problem statement, but with two different meanings. The point is just that the $\lim \sup$ always exists (if you allow $\pm \infty$ as possible values). But yes, I'm using $M_2$ to account for the possibility that early terms of the sequence have large values. Since there are only finitely many early terms, that's not a problem -- just take the largest. $\endgroup$ – Robert Shore Jun 20 at 0:49
  • $\begingroup$ I changed the designation of my sequence in the definition. I'm not sure if you can edit your comment, but if possible for completeness. Thank you $\endgroup$ – dc3rd Jun 20 at 0:55
  • $\begingroup$ Too late. There's only a limited window of time to edit comments. $\endgroup$ – Robert Shore Jun 20 at 0:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.